7.7: Limiting Reagent and Percent Yield

Introduction to Chemistry & the Environment · LibreTexts

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    7.7: Limiting Reagent and Percent Yield

    Which Reagent Runs Out First?

    By the end of this section, you will be able to:

    • Explain the concepts of theoretical yield and limiting reactants/reagents.
    • Derive the theoretical yield for a reaction under specified conditions.
    • Calculate the percent yield for a reaction.

    The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.

    Limiting Reagent

    Consider another food analogy, making grilled cheese sandwiches (Fig. 7.7.1)

    \[ \Large\boldsymbol{ 1\,\mathrm{slice\ of\ cheese} \quad + \quad 2\,\mathrm{slices\ of\ bread} \quad \longrightarrow \quad 1\,\mathrm{sandwich} } \]

    Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.

    Bread slices and cheese to show limiting reagent

    Figure \(\PageIndex{1}\): Using Cooking to Model Limiting Reagent. Cheese and bread slices are shown at the top and mimic reactants in a chemical reaction. To form sandwiches, they need to be put together in a 2:1 ratio (2 slices of bread for 1 slice of cheese). The first one to run out as you are making the sandwiches is the limiting reagent (ie. it limits how many sandwiches you can make!) In this case, the cheese runs out first and bread is left over. Cheese is, therefore, the limiting reagent.


    Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:

    \[ \Large\boldsymbol{ \mathrm{H}_{2}(g) \qquad + \qquad \mathrm{Cl}_{2}(g) \qquad \longrightarrow \qquad 2\,\mathrm{HCl}(g) } \]

    The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reagent, and the other substance is the excess reagent. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H2 and 2 moles of Cl2. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted.

    An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For example, if we had 3 mol H2 and excess Cl2, then we would produce 6 mol HCl and fully use up all of the H2 present:

    \[ \Large\boldsymbol{ 3\,\mathrm{mol}\,\mathrm{H}_{2} \qquad \times \qquad \frac{2\,\mathrm{mol}\,\mathrm{HCl}} {1\,\mathrm{mol}\,\mathrm{H}_{2}} \qquad = \qquad 6\,\mathrm{mol}\,\mathrm{HCl} } \]

    However, if we had 2 mol of Cl2 and excess H2, then we could produce a total of 4 mol of HCl to fully use up all of the Cl2 present:

    \[ \Large\boldsymbol{ 2\,\mathrm{mol}\,\mathrm{Cl}_{2} \qquad \times \qquad \frac{2\,\mathrm{mol}\,\mathrm{HCl}} {1\,\mathrm{mol}\,\mathrm{Cl}_{2}} \qquad = \qquad 4\,\mathrm{mol}\,\mathrm{HCl} } \]

    Therefore, when we have only 2 mol of Cl2 and 3 mol of H2, we can see that we will run out of Cl2 after 4 mol of HCl have been produced, and there will be leftover H2 gas that is unreacted. Thus, Cl2 is the limiting reagent because it runs out first, and limits the amount of product that can be formed.

    As seen above, the chlorine gas will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen gas was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine gas, therefore, is the limiting reactant and hydrogen gas is the excess reactant (Fig. 7.7.2).

    Hydrogen and chlorine interacting to form hydrochloric acid

    Figure \(\PageIndex{2}\): Finding the Limiting Reagent. When H2 and Cl2 are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant. Figure is from Wacowich-Sgarbi, S. and the Langara Chemistry Dept. (2018) BCCampus.


    View this interactive PHET simulation illustrating the concepts of limiting and excess reactants.

    Phet Simulation

    Example \(\PageIndex{1}\)

    Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

    \[ \Large\boldsymbol{ 3\,\mathrm{Si}(s) \qquad + \qquad 2\,\mathrm{N}_{2}(g) \qquad \longrightarrow \qquad \mathrm{Si}_{3}\mathrm{N}_{4}(s) } \]

    Which is the limiting reactant when 2.00 g of Si and 1.50 g of N2 react?

    Solution

    Step 1: Determine how many moles of Si are present.

    \[ \Large\boldsymbol{ 2.00\,\mathrm{g}\,\mathrm{Si} \qquad \times \qquad \frac{1\,\mathrm{mol}\,\mathrm{Si}} {28.09\,\mathrm{g}\,\mathrm{Si}} \qquad = \qquad 0.0712\,\mathrm{mol}\,\mathrm{Si} } \]

    Step 2: Determine how many moles of N₂ are present.

    \[ \Large\boldsymbol{ 1.50\,\mathrm{g}\,\mathrm{N}_{2} \qquad \times \qquad \frac{1\,\mathrm{mol}\,\mathrm{N}_{2}} {28.02\,\mathrm{g}\,\mathrm{N}_{2}} \qquad = \qquad 0.0535\,\mathrm{mol}\,\mathrm{N}_{2} } \]

    Step 3: Calculate how many moles of Si₃N₄ each reactant could produce.

    \[ \Large\boldsymbol{ 0.0712\,\mathrm{mol}\,\mathrm{Si} \qquad \times \qquad \frac{1\,\mathrm{mol}\,\mathrm{Si}_{3}\mathrm{N}_{4}} {3\,\mathrm{mol}\,\mathrm{Si}} \qquad = \qquad 0.0237\,\mathrm{mol}\,\mathrm{Si}_{3}\mathrm{N}_{4} } \]

    From the nitrogen:

    \[ \Large\boldsymbol{ 0.0535\,\mathrm{mol}\,\mathrm{N}_{2} \qquad \times \qquad \frac{1\,\mathrm{mol}\,\mathrm{Si}_{3}\mathrm{N}_{4}} {2\,\mathrm{mol}\,\mathrm{N}_{2}} \qquad = \qquad 0.0268\,\mathrm{mol}\,\mathrm{Si}_{3}\mathrm{N}_{4} } \]

    Conclusion

    Because silicon produces the smaller amount of product (0.0237 mol Si₃N₄), silicon is the limiting reactant.

    \[ \Large\boldsymbol{ \mathrm{Limiting\ Reactant} \qquad = \qquad \mathrm{Si} } \]

    Exercise \(\PageIndex{1}\)

    Which is the limiting reactant when 5.00 g of H2 and 10.0 g of O2 react and form water? (Hint: Start by creating the balanced equation).

    Answer

    O2


    Percent Yield

    The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete. Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield:

    \[ \Large\boldsymbol{ \mathrm{Percent\ Yield} \quad = \quad \frac{\mathrm{Actual\ Yield}} {\mathrm{Theoretical\ Yield}} \quad \times \quad 100\% } \]

    Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.

    Example \(\PageIndex{2}\)

    Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:

    \[ \Large\boldsymbol{ \mathrm{CuSO}_{4}(aq) \quad + \quad \mathrm{Zn}(s) \quad \longrightarrow \quad \mathrm{Cu}(s) \quad + \quad \mathrm{ZnSO}_{4}(aq) } \]

    What is the percent yield?

    Solution

    To find the theoretical yield of Cu, use the molar mass of CuSO₄ (159.61 g/mol) and the 1:1 mole ratio:

    \[ \Large\boldsymbol{ 1.274\,\mathrm{g}\,\mathrm{CuSO}_{4} \quad \times \quad \frac{1\,\mathrm{mol}\,\mathrm{CuSO}_{4}} {159.61\,\mathrm{g}\,\mathrm{CuSO}_{4}} \quad \times \quad \frac{1\,\mathrm{mol}\,\mathrm{Cu}} {1\,\mathrm{mol}\,\mathrm{CuSO}_{4}} \quad \times \quad \frac{63.55\,\mathrm{g}\,\mathrm{Cu}} {1\,\mathrm{mol}\,\mathrm{Cu}} \quad = \quad 0.5073\,\mathrm{g}\,\mathrm{Cu} } \]

    So the theoretical yield is 0.5073 g Cu.

    Then apply the percent-yield equation:

    \[ \Large\boldsymbol{ \mathrm{Percent\ Yield} \quad = \quad \frac{\mathrm{Actual\ Yield}} {\mathrm{Theoretical\ Yield}} \quad \times \quad 100\% } \]

    For this problem:

    \[ \Large\boldsymbol{ \mathrm{Percent\ Yield} \quad = \quad \frac{0.392\,\mathrm{g}\,\mathrm{Cu}} {0.5073\,\mathrm{g}\,\mathrm{Cu}} \quad \times \quad 100\% \quad = \quad 77.3\% } \]

    So the final answer is:

    \[ \Large\boldsymbol{ \mathrm{Percent\ Yield} \quad = \quad 77.3\% } \]

    Exercise \(\PageIndex{2}\)

    What is the percent yield of a reaction that produces 12.5 g of the gas Freon CF2Cl2 from 32.9 g of CCl4 and excess HF?

    \[ \Large\boldsymbol{ \mathrm{CCl}_{4} \qquad + \qquad 2\,\mathrm{HF} \qquad \longrightarrow \qquad \mathrm{CF}_{2}\mathrm{Cl}_{2} \qquad + \qquad 2\,\mathrm{HCl} } \]

    Answer

    48.3%

    Green Chemistry and Atom Economy

    The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the “Twelve Principles of Green Chemistry” (see details at this website). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:

    \[ \Large\boldsymbol{ \mathrm{Atom\ Economy} \quad = \quad \frac{\mathrm{Mass\ of\ Desired\ Product}} {\mathrm{Total\ Mass\ of\ Reactants}} \quad \times \quad 100\% } \]

    Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.

    The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Fig 7.7.3). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%.

    Production of Ibuprofen using Green Chemistry

    Figure \(\PageIndex{3}\): Green Chemistry and Ibuprofen Production. (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)Figure from: Wacowich-Sgarbi, S. and the Langara Chemistry Dept. (2018) BCCampusIn the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency’s Greener Synthetic Pathways Award in 1997.


    This page titled 7.7: Limiting Reagent and Percent Yield was last modified on Tue, 11 Aug 2026 18:40:33 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Hadeel Abozenadah, Adam Bishop, Scott Bitner, Kaysha Lehman and Patricia Flatt.